∵AD, DC, BG intercept ⊙ O at point A, point E and point B respectively.
∴DA=DE,CE=CB,
Let BC be X, then CF=x-2 and DC=x+2.
In Rt△DFC, (x+2)2-(x-2)2=(2 5)2,
Solution: x = 5/2;
∫AD∨BG,
∴∠DAE=∠EGC,
DA = DE,
∴∠dae=∠aed;
∠∠AED =∠CEG,
∴∠EGC=∠CEG,
∴CG=CE=CB= 5/2,
∴BG=5,
∴AG= (2 root number 5) 2+5 2 = root number 45=3 root number 5;
Solution 1: connect BE, S△ABG= 1/2AB? BG= 1/2AG? Yes,
∴ 2 5×5=3 5BE,
∴ BE= 10/3,
In Rt△BEG,
EG= radical number (BG 2-be 2) = radical number [5 2-( 10/3) 2] = 5/3 radical number 5,
Solution 2: 2:∠∠DAE =∠EGC, ∠AED=∠CEG,
∴△ADE∽△GCE,
∴ AD/CG=AE/EG, 5/2.5=3 radical 5-EG/EG, the solution is EG=5 radical 5/3.
Some 2' s are square, you can understand.